If $x \veebar y = 3x^{2}+y^{2}$ and $x \odot y = x(y-3)$, find $(-6 \odot 3) \veebar 3$.
Explanation: First, find $-6 \odot 3$ $ -6 \odot 3 = -6(3-3)$ $ \hphantom{-6 \odot 3} = 0$ Now, find $0 \veebar 3$ $ 0 \veebar 3 = 3(0^{2})+3^{2}$ $ \hphantom{0 \veebar 3} = 9$.